In an ap a 1 an 20 sn 399

Author: wfni

August undefined, 2024

WebIf an A.P. has a = 1, t n = 20 and s n = 399, then value of n is : A. 20 B. 32 C. 38 D. 40 Answer: Option C Solution (By Examveda Team) S n = 1 2 ( a + l) × n ⇒ 399 = ( 1 + 20) × n 2 ⇒ 399 × 2 = 21 × n ⇒ n = 399 × 2 21 ⇒ n = 19 × 2 ⇒ n = 38 Join The Discussion Related Questions on Progressions How many terms are there in 20, 25, 30 . . . . . . 140? WebFeb 22, 2013 · In an A.P. , if a = 1, an= 20 and sn =399, then n is? Share with your friends 1 Follow 0 Jai Prakash, added an answer, on 22/2/13 we know from that formula that, Sn = n/2 (a + an) 399 = n/2 (1 + 20) 399*2 = n (21) n = 798/21 = 38 Therefore n = 38 Was this answer helpful? 3 View Full Answer

In an AP, if a =1, an, = 20 and Sn = 399, then find n

WebAug 25, 2024 · In an AP, if a = 1, an = 20 and Sn = 399, then n is equal to A. 19 B. 21 C. 38 D. 42. ← Prev Question Next Question →. 0 votes. 17.2k views. asked Aug 25, 2024 in … greenish grey color name

In an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) …

WebMar 11, 2024 · answered In an Ap, a=1, an=20, sn=399 then find n Advertisement Answer 3 people found it helpful pragyanrana31 Answer: Sn=n/2 (a+l) . . 399=n/2 (1+20) . . 399*2/21=n . . 19*2=n . . N=38 . . Hope this helps you . . Pls mark Brainliest answer! ️ Find Math textbook solutions? Class 6 Class 5 Class 4 Class 3 Class 2 Class 1 NCERT Class 9 … WebJan 23, 2024 · In an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) 38 (D) 42. 2Pi classes. 36.7K subscribers. WebIn an A.P 1 st term is 1 and the last term is 20. The sum of all terms is =399 then n= .... A 42 B 38 C 21 D 19 Medium Solution Verified by Toppr Correct option is B) The is given that, First term (a)=1 Last term (t n)=20 Sum of terms (S n)=399 We know that, t n=a+(n−1)d S n= 2n(2a+(n−1)d) S n= 2n(a+(a+(n−1)d)) ⇒S n= 2n(a+t n) ⇒399= 2n(1+20) greenish grey couch

The first term of an A.P. is 1 and the nth term is20. If Sn = 399, …

In an AP, if d = -4, n = 7, an = 4, then a is, a. 6, b. 7, c. 20, d. 28

WebFeb 11, 2013 · if an ap a=1 ,an=20,sn= 399 then n is Asked by anish mishra 11 Feb, 2013, 02:45: PM Expert Answer Sn=399 => n/2 (a + an) = 399 => n/2 ( 1 + 20) = 399 =>n/2 (21) = 399 => n = 399 X 2/21 => n= 38 Answered by 11 Feb, 2013, 05:49: PM Application Videos This video explains the concept of Arithmetic Progression, Solving Linear E... WebIn an AP if a = 1, an = 20 and Sn = 399, then n is _____. CBSE English Medium Class 10. Question Papers 939. Textbook Solutions 33590. MCQ Online Mock Tests 12. Important … flyers christmas ornamentsWebFeb 27, 2012 · In an AP if a=1, an= 20 and sn =399 then n is a) 13 b) 38 c) 21 - Maths - Arithmetic Progressions - 1833845 Meritnation.com Class-10 » Maths Arithmetic … greenish grey colour

"WebAn arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous … " - In an ap a 1 an 20 sn 399

In an ap a 1 an 20 sn 399

In an AP: (i) given a = 8, an = 62. Sn = 210, find ‘n’ and ‘d’.

WebIf an AP has a=1, t n=20 and S n=399, then the value of n is A 19 B 32 C 38 D 40 Easy Solution Verified by Toppr Correct option is C) Given, a=1 t n=1+(n−1)d=20 ∴(n−1)d=19 S … WebLas protecciones eléctricas tienen un papel relevante en la seguridad y en la adecuada operación de un sistema eléctrico de potencia. Particularmente, en el caso de la generación y el trasporte, por su importancia requieren un eficiente sistema de protecciones que permita garantizar la integridad de sus elementos y la continuidad del servicio eléctrico.

Did you know?

WebFeb 27, 2012 · In an AP if a=1, an= 20 and sn =399 then n is a) 13 b) 38 c) 21 d)19 - Maths - Arithmetic Progressions WebFeb 14, 2024 · Solution For (2) In an AP if a=1,an =20, and Sn =399,0 then n is equal a=1an =20Sn =399

WebIf an A.P. has a = 1, t n = 20 and s n = 399, then value of n is : If an A.P. has a = 1, t. n. = 20 and s. n. = 399, then value of n is : A. 20. B. 32. C. 38. WebBesides giving the explanation of In an A.P. if a=1,An=20 and Sn=399, then find the value of n.?, a detailed solution for In an A.P. if a=1,An=20 and Sn=399, then find the value of n.? …

WebOct 25, 2015 · In an AP, if a=1,an=20 and Sn= 399, then n=? Advertisement Expert-Verified Answer 603 people found it helpful Brainly User since , an = a+ (n-1)d 20 = 1 + (n-1)d 19 = … WebIn an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) 38 (D) 42. The sum of first five multiples of 3 is ... Also find S 20. In an AP, if Sn = n (4n + 1), find the AP. 54 EXEMPLAR PROBLEMS. In an AP, if Sn = 3n 2 + 5n and ak = 164, find the value of k. If Sn denotes the sum of first n terms of an AP, prove that. S 12 = 3(S 8 –S 4 )

WebMar 28, 2024 · Objectives: This study aims to investigate the risk factors associated with severity and death from COVID-19 through a systematic review and meta-analysis of the published documents in Iran.Methods: A systematic search was performed based on all articles indexed in Scopus, Embase, Web of Science (WOS), PubMed, and Google Scholar …

WebSum of n terms of AP is given by: S n = n 2 2 a 1 + n-1 d where a 1 = first term, d = common difference. S 24-4 = 24-4 2 2 × 1 + 24-4-1 13 ∵ d = 13, n = 24, a 1 = 1 S 20 = 20 2 2 + 19 × 13 = 2490. a n-4 = a 1 + n-4-1 d a 24-4 = 1 + 24-4-1 13 ∵ d = 13, n = 24, a 1 = 1 a 20 = 248. Therefore, the ordered pair S n-4, a n-4 is equal to (2490 ... flyers chuck fletcherWebIn an AP if a = 1, a n = 20 and S n = 399, then n is 38. Explanation: ∵ S n = n 2 [ 2 a + ( n - 1) d] 339 = n 2 [ 2 × 1 + ( n - 1) d] 798 = 2 n + n ( n - 1) d ........ (i) And a n = 20 ⇒ a + ( n - 1) d = … flyers christmas packageWebIn an AP, if a = 1, an = 20 and Sn = 399, then n is equal to (a) 19 (b) 21 (c) 38 (d) 42 Solution: Question 18: The sum of first five multiples of 3 is (a) 45 (b) 55 (c) 65 (d) 75 Solution: (a) The first five multiples of 3 are 3, 6, 9,12 and 15. Here, first term, a = 3, common difference, d = 6-3 = 3 and number of terms, n = 5 flyer schul agWebHere d = 19 – 20 = (–1) and S n = 200 Hence, the required number of rows are 16 and in the top row there are 5 logs. 224 Views. Answer . The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP. Let the first term and the common difference of the A .P. be a and d ... flyers churchWebFeb 26, 2016 · WE HAVE- SN=399 , A=1 & AN=20. {Sn=n/2 (a+an)} =>so lets substitute... 399=n/2 (a+an) 399=n/2 (1+20) 399=n/2 (21) =>lets send 2 and 21 to RHS. 399X2/21=n. … flyers christmas partyWebQuestion If an AP has a = 1, a n = 20 and S n = 399, then find the value of n. Solution Compute the value of n: Formulae: The n th term of an AP is a n = a + n - 1 d and the sum of n terms of a series in an AP is S n = n 2 2 a + ( n - 1) d. Substitute a = 1, a n = 20 in the n th term of an AP formula. greenish gray tileWebIn an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) 38 (D) 42 Home Get Answers to all your Questions Home NCERT In an AP if a = 1, an = 20 and Sn = 399, then n is (A) 19 (B) 21 (C) 38 (D) 42 #Arithmetic Progressions #NCERT #Maths #Exemplar Maths for Class 10 #1.3 #5.2 #5.3 #5.4 #5.1 #9.2 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 greenish grey color